SUMMER - 2017 EXAMINATION Subject: Power System …msbte.engg-info.website/sites/default/files/s17mo s62a... · · 2017-07-02State the difference between ‘Generator bus’ and

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

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(ISO/IEC - 27001 - 2005 Certified)

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SUMMER - 2017 EXAMINATION

Subject: Power System Operation & Control Subject Code:

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17643

Important Instructions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the model

answer scheme.

2) The model answer and the answer written by candidate may vary but the examiner may tryto

assess the understanding level of the candidate.

3) The language errors such as grammatical, spelling errors should not be given more Importance

(Not applicable for subject English and Communication Skills).

4) While assessing figures, examiner may give credit for principal components indicated in the

figure. The figures drawn by candidate and model answer may vary. The examiner may give

credit for any equivalent figure drawn.

5) Credits may be given step wise for numerical problems. In some cases, the assumed constant

values may vary and there may be some difference in the candidate‟s answers and model

answer.

6) In case of some questions credit may be given by judgement on part of examiner of relevant

answer based on candidate‟s understanding.

7) For programming language papers, credit may be given to any other program based on

equivalent concept.

Q.

No

.

Sub

Q.N.

Answer Marking

Scheme

1. (A)

(a)

Ans.

Attempt any THREE:

State the difference between ‘Generator bus’ and ‘slack bus’.

Sr.

No

Generator Bus Slack Bus

1. At this bus power

generated is injected into

the system

Oneof the generator bus is

made to take additional

real and reactive power to

supply transmission losses.

This bus is known as slack

or swing bus.

2. Specified quantities in

this bus is P & V.

Specified quantities in this bus

is V &.

3. Unknown quantities in

this bus- Q &.

Unknown quantities in this

bus- P & Q.

4. It is not a reference bus. It is a reference bus.

5. This bus does not

provides additional real

& reactive power losses

This bus provides additional

real & reactive power losses

12

4M

Any 4

point

1M for

each


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(b)

Ans.

State the concept of reactive power compensation. Name any two

reactive power compensating equipments.

concept of reactive power compensation: The main objective of the

utilities is to satisfy the consumers with its power demand. To meet

the consumer‟s reactive power demands, if the same power is

generated at generating stations & feed to the consumer, it will cause

voltage drop in line. This will result into reduction of transmission

efficiency and the cost of power transmission increases. Instead of

this is we generate power locally near the load centers& feed it to

consumers to his satisfaction the performance of power system will

not affect & cost of power transmission also will not increase.

Reactive power generating equipments are located near the load

centers which will help to meet the reactive power demand of

consumers to his satisfaction. These also help to control the voltage

levels in the system. The methods used for this is also called as

“Reactive Power Compensation “. And the equipment used is called

as “reactive power compensating equipment”. Reactive power

compensating equipment can be employed either at load level,

substation level, or at transmission level.

Reactive power compensating equipments:-

1. Shunt compensation equipment‟s - Shunt reactor, shunt

capacitor

2. static VAR system

3. Series compensation equipments - Series reactors

4. Synchronous compensation equipments - Synchronous

condenser

4M

2M for

concept

2M for

equipme

nts (any

2)

(c)

Ans.

Define the following terms.

- Power system stability.

- Power system instability.

- Power system stability limit.

Power System Stability: It is the ability of power system to return to normal or stable

operation after having been subjected to some form of disturbance.

4M

Power

system

stability

1M


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Power System instability: It is status of system when it loses its

normal stable operating condition because of sudden increase/

decrease in power demand or due to occurrence of major fault.

Power system stability limit:

The maximum feasible power flow through some point in the power

system. When the entire system or port thereof under investigation is

operating with stability.

Power

system

instabili

ty &

Stability

limi 1½

M each

(d)

Ans. State the adverse effects of instability on power system.

Due to instability of power system following effects can be observed:

1) As the Pd>>Pg, frequency of system varies over a wider range/

beyond the limits. Hence protective scheme of generators,

transformersmaytrip them.

2) Due to fluctuation of V, F, P, Q performance of grid network

reduces and power transmission capacity reduces and so consumers

receive poor quality of supply.

3) If the system parameters are not controlled, than one by one

generator will trip off and it leads major failure sometimes leads to

collapsing of whole system.

4) As consumer receives poor quality supply the performance of their

machines reduces, production rate decreases, quality of product

reduces and overall there is financial loss.

4M

Each

effect

1M

1. (B)

(a)

Ans.

Attempt any ONE:

State the importance of load flow analysis in operation of power

system.

Following are the reasons that shows importance of load flow

analysis in operation of power system:

1) The total amount of real power flow thro‟ the network generates at

generating stations whose size and location are fixed.

2) At each moment power generation must be equal to power demand

as electrical power cannot be stored.

3) Hence the load on the system has to be divided between no. of

generators in a unique ratio in order to achieve optimum economic

power generation.

4) Hence the generator output must be closely maintained at

06

6M

1M each

point


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predetermined set closely maintained at predetermined set points.

5) It is important to remember that the power demand under goes

slow but wide variation throughout the 24 hrs. of the day.

6) Therefore we must change these set points slowly or continuously

or in discrete step as the hours wear on. This means that load flow

configuration that fits the demand of a certain hr. of the day may

look quite different next hour.

(b)

Ans. List out the significant features of Ybus matrix.

Features of Ybus:

1) Ybusis a symmetrical matrix “n x n” matrix.

2) All diagonal elements Yii represent “self-admittances‟ of bus “I”.

3) All off diagonal elements Yij represents mutual admittance

between bus “I” bus “j”.

4) With reference to mutual admittance

Yij= Yjii.e. Y12 = Y21, Y13 = Y31

Hence it is a symmetricalmatrix.

5) Any element in the matrix “zero‟ indicates that there is not to line

between those buses.

Y21 = Y12 = 0 ………… No tr. line between bus I bus II or outage

of tr. Line

Yik = Yki = 0 if i k between but I bus II i k are not connected.

6) Ybus = (Zbus)-1

whereZbus– bus impedance matrix.

7) All elements are complex numbers.

8) Self admittances are defined as Y11 = y 11 + y 12+ y 13

Where y11 – line changing admittance

y12, y13 – line admittances

Y11 = sum of line changing admittance and total line admittances

connected to a bus.

9) Mutual admittances are defined as

Y12 = -y 12 = -y 21 = Y21 and Y13 = -y 13 = Y31

i.e. mutualadmittance is negative of line admittance between two

buses.

10) All mutual admittances are negative complex numbers.

6M

Any 6

1M each

2.

(a)

Ans.

Attempt any FOUR:

Why power utilities maintain the fluctuation in frequency within

the tolerance limit? Power utilities maintain the fluctuation in frequency within the

tolerance limit because:

1. In thermal power station steam turbines are especially designed to operate

16

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at a very precise speed for required efficiency. The velocity of the

expanding steam is beyond our control. Hence variation in frequency

results in reduction in life of the turbine and generator.

2. In hydro power stations a turbo rotor with its many large turbine blades

consists a mechanical system of many natural frequencies. These

frequencies are quite un-damped and at various speeds they are subjected

to resonance. Hence it is very important that under load condition rotor

should never drift into such a speed range where dangerous amplitudes of

blades are build up. Hydro turbines are not supposed to subject to this

dangerous condition.

3. The overall operation of power system can be much better controlled if

the frequency error is kept within strict limits. By reducing normal

frequency fluctuations to a faint ripple, fault can be detected at early

stage. Hence in modern power system the frequency variation is normally

kept within + 0.05 Hz

4. Power systems are interconnected through H.V. line to meet increase in

demand. Hence to regulate the power flow through these lines, need

accurate constant frequency.

5. The operation of a transformer below the rated frequency is not desirable.

When frequency goes below rated frequency at constant system voltage

then the flux in the core increases and then the transformer core goes into

the saturation region.

6. With reduced frequency the blast by ID fans and FD fans decrease, and so

the generation decreases and thus it becomes a multiplying effect and may

result in shut down of the plant.

Any 4

point

1M each

(b)

Ans.

Derive the relation between real power and frequency

considering a simple two bus system.

Consider a simple two bus system in which power is transmitted from

bus 1 to bus 2 through a short transmission line.

Let 𝑉11be the voltage at bus 1,

𝑉22 be the voltage at bus 2

Z be the total series impedance of the transmission line per phase =

R+jX

4M

Diagra

m 1M

Explana

tion 3M


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Since R<<<X, Z=jX

I be the current through the line per phase

I= 𝑉11−𝑉22

𝑧 =

𝑣1𝐿𝛿1−𝑣2𝐿𝛿2

𝑗𝑋

S12 - complex power transferred from bus-1 to bus-2 = V1 I1*

=V12

- V1 V2(δ1 - δ2) x j X = jXV12

- j X V1 V2(δ1 - δ2)

-j X j X X2

= jV12 - j V1V2(δ1 - δ2)

X

= jV12 - j V1V2 ( cos δ –j sin δ) where δ1 - δ2 = δ

X

= V1V2 Sin δ + j (V12 – V1V2 Cos δ) = P1 +j Q1

X X

So now, P1 = V1V2 Sin δ ----------------------------------1

X

Q1 = (V12 – V1V2 Cos δ) ------------------------2

X

Similarly, we can also calculate complex power S21 that flows from

bus-2 to bus-1.

We have,

S21 = V2 I2* = V2δ2 ( V2-δ2 -- V1-δ1 )

j X

= V1V2 Sin δ + j (V1V2 Cos δ– V22) = P2 +j Q2

X X

Therefore,

P2 = V1V2 Sin δ --------------------------------3

X

Q2 = ( V1V2 Cos δ -V12) -----------------------4

X

From equations 1 & 3, we get……

P1 = P2 = V1V2 Sin δ

X

Therefore, net power flow through the line is…..

P= V1V2 Sin δ

X


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For given system X is constant. If voltages at both ends of the line are

maintained constant i.e. V1& V2 remains same.

Then, 𝑃 ∝ 𝑆𝑖𝑛𝑒𝛿 ……………Where δ = ωt=2∏ƒt

𝛿 ∝ 𝑡 This shows that variation in real power flow results in variation of

supply frequency. P is maximum when Sin δ = 1

P = Pmax when δ = 900

But to operate the power system under stability condition the value δ

is maintained between 350 to 45

0.

(c)

Ans.

State the data required for load flow, analysis about-

Transformer, Triline, Generator and Bus.

1. Transformer data: Type of transformer such as distribution

transformer / power transformer, auto-transformer, tap-changing

transformer (on-line or off-line).

Also ratings, % impedance and tap setting points, tap setting on HV

/LV /both sides are required. Resistance of the transformer,

reactance of the transformer, and the off nominal turns-ratio.

provision for tapping‟s

For every transformer connected between buses iandk the data to

be given includes: the starting bus number i, ending bus number k,

2. Transmission line data: For every transmission line connected

between buses I and k the data includes the starting bus number i,

ending bus number k,

- Line parameters – .resistance of the line, reactance of the line

and the half line charging admittance. Series impedance (z) in

per unit, shunt admittance (Y) in per units,

- Thermal limits of the line.

- Length of the line. SG1=PG1 + j QG1

- Identification of each line and its „∏‟ equivalent circuit.

3. Generator data: No. of generating stations connected in the

systemready to generate the required amount of power and their

time duration should be available. Each generators rating,

maximum& minimum limits of generation, their characteristics, and

excitation control details are made available.

4.Bus data: Depending upon no. of buses in the system, bus data

4M

1M for

each

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should be made available.

Type of bus Bus data No. of

buses

For each Bus

Generator bus P, (V) P, Vi, , minimum

and maximum

reactive power

limit (Qi,min,

Qi,max).

Load bus P, Q Active power

demand PDi, and

the reactive power

demand QDi.

Slack bus V, δ Generator ratings

which is assume to

be connected to

slack bus

Voltage

control bus

P Q V Voltage control

equipment used

and its rating, max.

& min. limits

(d)

Ans. Write general form of SLFE considering two-bus system.

For a two bus power system , Load flow equations can be written

as……

𝑉1𝛿1 be the voltage at bus 1, 𝑉2𝛿2 be the voltage at bus 2

Y11– self admittance of bus-1 Y22– self admittance of bus-2

Y12 = Y21– mutual admittance between bus-1&bus-2

Then real power at bus -1…

P1 = PG1 – PD1= V1 2

Y11 Cos α11 +Y12 V2 V1 Cos ( δ2 –δ1 + α12)

Reactive power at bus -1…

Q1 = QG1 – QD1= – [ V1 2 Y11 Sin α11 +Y12 V2 V1 Sin ( δ2 –δ1 +

α12)] Real power at bus -2…

P2 = PG2 – PD2 = Y21 V2 V1 Cos ( δ1 –δ2 + α21)+V2 2 Y22 Cos

α22 Reactive power at bus -2….

Q2 =QG2 – QD2 = – [ Y21 V2 V1 Sin ( δ1 –δ2 + α21)+V2 2 Y22 Sin

α22]

1M for

each

equatio

n


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OR

Consider a power system having „n‟ no. of buses,

and𝑉1𝛿1 ,𝑉2𝛿2 ,𝑉3𝛿3 ,…..………….𝑉𝐾𝛿𝐾……….𝑉𝑛𝛿𝑛 are the bus

voltages.

Let Y11α11, Y22α22,Y33α33,……….Ykkαkk……….Ynnαnn are self-

admittances of bus1,2,3….k….n resply.

Let Y12α12, Y13α13,Y14α14,……….Y1kα1k……….Y1nα1n are mutual

admittances of bus-1 with bus,2,3….k….n resply.

Similarly…

Y21α21, Y23α23,Y24α24,……….Y2kα2k……….Y2nα2n are mutual

admittances of bus-2 with bus,1 ,3….k….n resply.

And similarly for all buses.

Now SLFE can be written as…….

Power at 1st bus can be written as….

P1 = V1 2 Y11 Cos α11 +Y12 V2 V1 Cos (δ2 –δ1 + α12) + Y13 V3V1 Cos (δ3 –δ1 +

α13)+ …...… ………………………...+ Y1n VnV1 Cos (δn–δ1+ α1n)

Q1 = -- [ ( V1 2

Y11 Sin α11+Y12 V2 V1 Sin (δ2 –δ1 + α12)+ Y13 V3V1 Sin

(δ3 –δ1 + α13)+……

………………..+ Y1n VnV1 Sin (δn–δ1+ α1n) ]

Power at k th

bus can be written as….

Pk= Yk1 V1VkCos (δ1 –δk+αk1)+Yk2 V2 VkCos (δ2 –δk+αk2) + Yk3 V3Vk

Cos (δ3 –δk+αk3) +…….+ Vk2

YkkCosαkk+……… …………………+

YknVnVk Cos (δn–δk +αkn)

Qk= --{ Yk1 V1VkSin (δ1 –δk+αk1)+Yk2 V2 VkSin (δ2 –δk+αk2) + Yk3

V3VkSin (δ3 –δk+αk3) +…….+ Vk2

YkkSinαkk+………

…………………+ YknVnVkSin (δn–δk +αkn)}

(e)

Ans.

State and explain Bus-loading and Line-flow equations refer to

power system.

Bus Loading:

The real of reactive power at any kth

bus can be written as

Sk = Pk – jQk = Ik

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Current at bus i IK = Pk–jQ K

𝑉𝑘∗

Bus current Ik is positive when if flows into the system.

If the shunt elements are neglected or considered in Ybus matrix then

above equation represents bus current i.e. bus loading Ik = Ikk If the

shunt elements is considered then equivalent ckt. For bus can be

drawn as.

Now

Now the net current through bus - Ikk = Ik - Ikg = - VkYkk When Ykk –

total shunt admittance

Tr. Line Flow equation:

Assume that current is flowing throl tr. Line from bus k to bus j.

Ikj= 𝑉𝑘−𝑉𝑗

𝑌𝑌1 + 𝑉𝑘 𝑌 2

Where Y1– line admittance Y – line changing admittance Now power

Explana

tion of

bus

loading

2M

Line

flow

equatio

n 2M


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flow from bus k to bus j is Pkj-jQkj = Similarly power flow from bus j

to bus k is Pjk – jQjk = The above two equations are called as “Line

flow equation”. The algebraic sum of power expressed by above

equations gives power loss in the transmission line k – j

(f)

Ans.

Develop a Ybus matrix for the following 3-bus system.

4M

Each

step 1M


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3.

(a)

Ans.

Attempt any FOUR:

State the comparison between shunt and series compensating

equipments. (any four factors).

Sr.

No

Shunt Compensating

Equipment

Series Compensating

Equipment

1. A device that is

connected in parallel

with a transmission line

is called a shunt

compensation equipment

A device that is connected in

series with the transmission line

is called a series compensation

equipment

2. the shunt compensator is

always connected at the

midpoint of transmission

system

he series compensator can be

connected at any point in the

line.

3. Shunt Reactive Power

compensation equipment

- Shunt reactor , shunt

capacitor

Series Reactive Power

compensation equipment -

Series reactors, Series

capacitors

4. shunt compensation is to

improve voltage profile.

series compensation is for

system stability.

16

4M

1M each

point

(b)

Ans. Why consumers demand constant voltage supply?

Consumer demand constant voltage supply because:

1. Due to variation in the supply voltage the current drawn by the

equipment varies.

2. When supply voltage decreases beyond the limit the current drawn

by equipment increases & efficiency decreases. As a result

performance of the equipment also reduces its life.

3. The induction motor which is commonly used as industrial drive

develops the torque which depends on supply voltage T 𝑉2.

Hence large variation in supply voltage leads to more variation in

torque developed .So far small variation in supply voltage the

performance of motor gets affected and as a result the quality of

product& the process gets affected.

4. In the lighting system the luminous output of lamp sources depends

on supply voltage. Light flux of a lamp depends on voltage, with

the voltage fluctuation light flux varies strongly As supply voltage

decreases the luminous output of lamp decreases with more

fluctuation in supply voltage reduces life of lamp also reduces.

4M

Any 4

point

1M each


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5.Nowadays the more sophisticated equipment are used e.g

computers which are very much sensitive towards supply

parameters. Fluctuation in supply voltages damages these

instruments permanently.

Because of above reason consumer demand constant supply.

(c)

Ans.

List out the informations that can be collected from load flow

studies.

(1) We get MW and MVAR flow in the various parts of the system

network.

(2) We get information about voltages at various buses in the system.

(3) We get information about optional load distribution.

(4) Impact of any change in generation (increase or decrease) on the

system.

(5) Influence of any modification or extension of the existing circuits

on the system loading.

(6) It also gives information for choice of appropriate rating and tap-

setting of the power transformer in the system.

(7) Influence of any change in conductor size and system voltage

level on power flow.

4M

Any

four

points

1M each

(d)

Ans. State the significance of Ybus matrix in loud flow studies.

Significance of Ybus matrix in load flow studies:

1. Y bus is a nodal matrix , is called as a nn matrix which describes

power system network which having n number of buses .

2. Admittance matrix is use to analyze data which is use in power

flow study.

3. Y bus matrix explain admittance & topology of network.

4. Outage of line or outage of generator can be easily indicated

through Y bus matrix in load flow analysis.

5. The off diagonal elements of matrix indicate mutual admittance i.e

the transmission line admittance between 2 bus.

6. All diagonal element indicate the sum of line charging admittance

& transmission line admittance. It is also called as self admittance

of bus.

4M

Any 4

Points

1M for

each

(e)

Ans.

When we can say that the power system is in ‘transient stability

condition’ or in steady state stability condition?

Transient state stability condition:

1.Transient state stability is the ability of the system to return to its

normal operating condition of same equilibrium or new equilibrium

position after experience sudden and large disturbance in the network.

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2. Large disturbance is the cause of Transient state stability. Large

disturbance means occurrence of fault, sudden increase in large

amount of load, switching operation etc.

Steady state stability condition:

1. A power system is in steady state operating condition if all the

measured or calculated physical quantities describing the operating

condition of the system can be considered as constant for purpose of

analysis.

2. Small disturbance is the cause of Steady state stability . Small

disturbance is nothing but change in setting of automatic voltage

regulator of excitation system.

Stable operation of system is obtained when >0 &<90o. But

practically it is maintained between 35o-45

o. If load angle delta falls

below 10o then system drawn into transient stability condition

2M for

each

point

(f)

Ans.

List out the factors that affects the transient stability condition of

a power system.

Following are the factors that affects the transient stability

condition of a power system:

i) Generators play a vital role in any power system. Their

characteristics have a significant impact on the stability

characteristics of the system.

ii) Under transient conditions, the transient reactance, rotor inertia

(inertia constant), excitation response and the electrical damping

provided by the generator rotor and the mechanical damping by the

prime mover determine the generator performance.

iii)From swing equation the acceleration of the rotor d2/dt

2 is

inversely proportional to the moment of inertia of the machine,

when accelerating power is constant, which means higher the

moment of inertia, the slower will be the change in rotor angle,

hence longer time for breaker operation.

iv)By reducing the switching time and also the transient reactance,

power limit can be substantially improved.

v)Voltage regulators improve stability limits subsequent to the first

swing oscillation, after the clearing of the faulty section.

vi) Excitation response

4M

Any

four

points

1M each


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4. (A)

(a)

Ans.

Attempt any THREE:

Draw neat labeled schematic diagram of a ‘turbine speed

governing system’.

12

4M

Correct

diagram

with

labeled

4M

(b)

Ans.

Explain the functioning of each component of ‘turbine speed

governing system’.

TheTurbine speed Governing system consists of the following

components:

i) Speed Changer:

It provides a steady state power output setting for the turbine. Their

downward movements open the upper pilot valve so that more steam

is admitted to the

ii) Fly ball speed governor:

This is the heart of the system which senses the change in speed. As

the speed increases the fly balls move outwards and the point B on

linkage mechanism moves downwards. The reverse happens when the

speed decreases.

iii) Linkage mechanism:

ABC is a rigid link pivoted at B and CDE is another rigid link pivoted

4M

Each

compon

ent 1M


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at D. This link mechanism provides a movement to control valve in

proportion to change in speed. It also provides a feedback from the

steam valve movement (link 4)

iv) Hydraulic amplifier:

It comprises a pilot valve and main piston arrangement. Low power

level pilot valve movement is converted into high power level piston

valve movement. This is necessary in order to open or close the steam

valve against high pressure steam.

Speed changer moves downward to raise the speed, fly ball of speed

governor moves outward, ABC link moves upward so that CDE link

moves downward. Now pilot piston of Hydraulic amplifier moves

downward and more oil rushes in and pushes the main piston in

downward direction. As the main piston move downward the valve

opens further and more steam in rushes to enter turbine. Now kinetic

energy increases and speed of turbine increases which will increase

the generator speed and generator output.

Whenever generator output has to decrease vise versa happens.

(c)

Ans.

Draw a neat labelled following curves and write the expressions.

- Input output curve

- Incremental fuel cost curve.

Input output curve:

𝑑𝐹

𝑑𝑃= 𝐹𝑛𝑛 Pn + fn = λ𝐧

4M

Input

output

curve

2M


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Incremental fuelcost curve:

𝑑𝐹

𝑑𝑃= 𝐹𝑛𝑛 Pn + fn = λ𝐧

Increme

ntal fuel

cost

curve

2M

(d)

Ans.

With reference to Indian Power system, state the types of LDCs

and their locations.

Types of LDC:

For increasing the reliability of supply, security of the system and for

increasing the stability limit of the system Load Dispatch

Centers(LDC) are located. In India Government has identified

different types of LDC depending on their locations.

National Load Dispatch Centre (NLDS)

Regional Load Dispatch Centre(RLDS)

State level Load Dispatch Centre(SLDS)

Location :

1) In India, there is one NRLDC located at Delhi.

2) In India each state has its own SLDS and in Maharashtra SLDS is

located in Nagpur.

3) In India there are four RLDS – Western, Eastern, Northern and

Southern. Western RLDS is located in Kalwa.

4)

RLDCs: There are region wise 5 RLDCS and regions are

Name of the

Region

States

included

Name of

RLDC

Located

NORTHERN

Region

Jammu and

Kashmir, Uttar

NRLDC Delhi

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Pradesh,

Punjab,

Hariyana

EASTERN

Region

Bengal, Bihar,

Orissa

ERLDC Kolkata

SOUTHERN

Region

Karnataka,

Goa,

Kerala,Tamiln

adu

SRLDC Banglore

WESTERN

Region

Mharastra,

Gujarat,

Madhya

Pradesh,Chatti

s gad

WRLDC Mumbai

NORTH-

EASTERN

Region

Assam, Sikim,

Nagaland

NERLDC Shillong

4. (B)

(a)

Ans.

Attempt any ONE:

State and explain any three conventional techniques used to

improve transient stability condition.

Conventional /Traditional Technique:

i) Effect of generator design.

ii) Increase of voltage level

iii) Reduction in transfer reactance

iv) Rapid fault clearing

v) Automatic reclosing of CB

Following are conventional methods adopted to improve transient

stability condition of a power system…..

i.) Effects of Generator Design:A heavy machine hasgreaterinertia

and is more stable than a light machine. Modern machines are

designed to get more power from smaller machines but this is

undesirable from the stability point of view. In earlier days a large

number of machines were employed to generate more power and this

is also not desirable from stability point of view. Salient pole

alternators operate at lower load angles and hence they are more

preferred than cylindrical rotor generates from considerations of

stability.

ii.) Increase of voltage: The amplitude of the power angle curve is

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directly proportional to the internal voltage of the machine.An

increase in voltage increases the stability limit.

iii.) Reduction in transfer reactance: The amplitude of the power

angle curve is inversely proportional to the transfer reactance. This

reactance can be reduced by connecting more line in parallel.

iv.) When two lines are connected in paralleland a fault occurs in one

line then some power is transferred to healthy line (except when the

fault is at receiving end or sending end bus). This transmission of

power helps the stability of the system.

v.) Some features of the power system layout and business

arrangement also help in improving stability.

vi.) Use of bundled conductors helps in reducing line reactance and

improving line stability.

vii.) The compensation of line reactance by series capacitance is

another effective method of improving stability.

viii.) Rapid fault clearing: By decreasing the fault cleaning angle (by

using high speed breakers) stability can be improved.

ix.) Automatic Reclosing: Most of the fault‟s on the transmission

lines are of transient nature and are self-clearing. Modern circuit

breakers are mostly of reclosing type. When a fault occurs, the faulted

line is de-energized to suppress the are in the fault and then the

circuit breaker recloses, after a suitable time interval.

(b)

Ans.

State and explain any four planning tools used for load

forecasting in power system operation.

Types of planning tools:

i. Simulation Tools: Load flow models, sc models transient stability

models, production costing, adequacy calculations.

ii. Optimization tools: Optimum power, least cost expansion

planning, generating expansion planning

iii. The scenario techniques: Sequence of events recording, possible

outcomes, decisions, assumptions, computerize and automatic

system.

i. Stimulation Tool: This tool help stimulate the behaviour of the system under certain

load condition. This helps to calculate certain relevant indices. i .e

cost of generation , transmission & distribution. Corporate models

simulate the impact of various decision of financial performance of

utilities. It requires voluminous data and required result from various

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models to be integrated.

Eg: Load flow model, short circuit model, Transient stability model,

production costing, estimation of environmental impact. Results

obtained are reliable as we wouldn‟t experience major failure.

ii. Optimization Tools: This tool minimizes or maximizes adequate

values for decision variables. Eg: Optimum power, least cost

expansion planning of generation. For example, we considered the

expansion of transmission circuit and planning for electrification rural

areas. Though the cost involved is very high, still we can implement

it, because objective behind it is on higher side (Socio economic

harnessing of ground water resource food production rural

employment prevention of migration).

iii. Scenario Tool: This tool is used to known the future in quantitative fashion. In this

technique narrative description is developed which includes probable,

sequential or simultaneous recorded data. This can be built up into

case history. A decision points are always identified and possible

outcomes are investigated. The sort of decision or assumption made

by utility is noted. All these narrative descriptions are computerized

and used as past data. After certain period it is also used in

“automatic power management” as data. Electrical utilities should

prepare integrated resource plan. This long term plan must develop

the best mix of demand and supply options to meet consumers need.

5.

(a)

Ans.

Attempt any FOUR:

Derive the expression for max steady state power in a simple two

bus system. Neglect the losses in the system.

This can be further simplified by considering single machine

connected to an infinite bus

Let, V1 = V1∟𝛿2 be the voltage of infinite bus.

The equivalent ckt. of the above system can be written as

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Let Xa * Xl are the reactance of generator & tr. Line.

Consider losses are neglible∴ R = 0.

Now the complex power injected by generator into the system is

OR

The variation of P w.r.t. 𝛿 can be represented as

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P = V1V2 sin δ

X is called “power angle equation” and curve is called

“power angle curve” or “power angle diagram”.

Consider two machine system as shown by single line diagram.


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(b)

Ans. Write swing equation and state significance of power angle.

“swing equation” which is written as 𝑀𝑑2𝛿

𝑑𝑡= 𝑃𝑎 = 𝑃𝑚 − 𝑃𝑒 = Ta = Tm –Te

where, M = I ω- angular momentum and I- Inertia constant

Pa = Ta ω -- Accelerating Power

Pm = Tm ω -- Mechanical power input to synchronous

generator

Pe = Te ω -- Electrical Power output of synchronous generator

= Ɵ ω t in radian -- rotor angular displacement.

Significance of power angle

P =V1V2

Xsinδ

1) As varies, power flow through the system also varies though

voltage profile at each end is maintained constant.

2 ) when =90 , power flow is max . When =0 , power flow is zero

3) when > 90 , system will be unstable

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(c)

Ans.

Draw a schematic diagram of Automatic voltage control at

generators. Explain its functioning.

Automatic Voltage Control/Automatic Voltage regulator (AVR):

The automatic voltage, regulator (AVR) loop controls the magnitude

of the terminal voltage V. The latter voltage is continuously sensed,

rectified and smoothed. This D.C. signal, the resulting „error

voltage‟, after amplification and signal shaping serves as the input to

the exciter which finally delivers the voltage Vf to the generator field

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winding.

What is the function of AVR? The basic role of the AVR is to provide reliability of the generator terminal voltage during normal, small and slow changes in the load. The main objective of Automatic Voltage Controller is to control of excitation of generator with respect to change in generated voltage. Automatic Voltage regulator is a feedback loop system, consists of voltage sensor, Automatic voltage regulator and Excitation system.

Voltage Sensor: It senses the actual generated voltage and sends the

signal to rectifier.

Rectifier: Now the rectifier converts into an equivalent D.C. Voltage-

Vdc.

Comparator: This D.C. voltage is fed to the comparator where it is compared with reference voltage Vdc ref.. The reference voltage is D.C. equivalent of the specific voltage required to maintain at generator output to maintain stability of system at that moment. The value of this is obtained by Load Flow Analysis.

Amplifier &Q-V controller: The output of comparator∆V is amplified by Power amplifier and amplified signal is given as input to Q-V controller. NowQ-V controller transfers into reactive power

signal ∆Q and feeds it to excitation controller. Excitation Controller:

This in turn varies the field regulator so that the generator terminal voltage varies.

(d)

Ans. Explain the load frequency control refer to single area case.

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In power system network single area is identified as single control

area of the grid network, consisting number of generators supply

power to all consumers in that area. All generators in this area are

synchronized and they swing in unison to meet change in power

demand of that area. i. e. they speed up or speed down

simultaneously.

Consider a control area consisting two generating plants connected

through transmission line as shown in single line diagram. These

generators are running in synchronism and their speed varies together

by maintain their respective machine angels.

At steady state condition, PG1 + PG2 = PD1 + PD2.

As load on generator G2 increases, its output increases to meet the

demand, while output of G1 remains same. But speed of G1 varies to

maintain the synchronism and frequency.

As load on generator G2 go on increasing, its output increases up to

its upper generation limit. Now to meet further increase in power

demand, output of G1 will increases and shares the power demand. If

the demand is very large then both generators share the demand.

Explana

tion 3M

(e)

Ans. State the need of load forecasting for power system operation.

Need of Load forecasting : Load forecasting for power system operation is required

1. For proper planning, designing of new power system network or

expansion of existing.

2. For varying generation of power with respect to time i.e. amount

of growth expected in power demand.

3. For determining the capacity of power generation and power flow

through transmission lines and distribution lines.

4. For proper planning of power flow through transmission &

distribution network.

5. For proper operation of power system in instability condition.

6. For proper planning of resources for generation of power i.e.

conventional or non-conventional resources.

7. For determining the cost of power generation, transmission and

distribution.

8. For proper man power development or training of manpower for

operation of power system.

9. To decide power tariff for different utilities and different

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consumers.

10. For proper power transaction between neighbouring grid system.

11. For proper energy sales in electrical market.

12. For finding the requirement of fuel in future.

(f)

Ans. List out the functions of LDC. (Any four)

Functions of State Load Dispatch Centers:

In accordance with section 32 of Electricity Act, 2003 roles and

functions of SLDCs are as under:

1. The State Load Dispatch Centre shall be the apex body to ensure

integrated operation of the power system in a State.

2. The State Load Dispatch Centre shall -

a. be responsible for optimum scheduling and dispatch of

electricity within a State, in accordance with the contracts

entered into with the licensees or the generating companies

operating in that State;

b. monitor grid operations;

c. keep accounts of the quantity of electricity transmitted

through the State grid;

d. exercise supervision and control over the intra-state

transmission system; and

e. be responsible for carrying out real time operations for grid

control and dispatch of electricity within the State through

secure and economic operation of the State grid in accordance

with the Grid Standards and the State Grid Code.

3. The State Load Dispatch Centre may collect such fee and charges

from the generating companies and licensees engaged in intra-

State transmission of electricity as may be specified by the State

Commission.

4. Overall supervision, monitoring and control of the integrated

power system in the State on real time basis for ensuring stability,

security and economy operation of the power system in the State.

5. Optimum scheduling and dispatch of electricity within the State.

For this SLDCs estimate the demand of the State / DISCOMS, as

may be the case, availability of power in the State/DISCOMS

from State generators and other sources like Central Generating

stations, bilateral contracts etc, conveys the final requisition to

RLDCs on the State‟s entitlement from the Central Generating

Stations and bilateral transactions under open access, if any, and

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issues final dispatch schedule to the State Generators and drawal

schedule to the DISCOMS.

6.

(a)

Ans.

Attempt any FOUR:

State the advantages of ALFC and AVC systems.

i. Automatic Load Frequency Control (ALFC) or Automatic

Generation Control (AGC) - Used to achieve real power balance

(acceptable frequency values).

ii. Automatic Voltage Regulator (AVC).--- Used to achieve reactive

power balance (acceptable voltage profile).

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advanta

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(b)

Ans.

Write the help of diagram, explain ALFC of synchronous

generator.

In an electric power system, Automatic Generation Control (AGC) is

a system for adjusting the power output of multiple generators at

different power plants, in response to changes in the load. Ideally it is

required that at each moment in power grid power generation and

power demand closely balance. To achieve this frequent adjustments

are necessary to the output of generators. The imbalance can be found

out by measuring the system frequency. If the system frequency is

increasing, means more power is being generated than demand and all

the machines in the system are accelerating. If the system frequency

is decreasing, means more loads is on the system, than all the

machines in the system are slowing down.

The frequency is closely related to the real power balance in the

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https://en.wikipedia.org/wiki/Electrical_grid

https://en.wikipedia.org/wiki/Power_station

https://en.wikipedia.org/wiki/Utility_frequency


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overall network. Under normal operating conditions the system

generators run synchronously and generate together the power at each

moment is being drawn by all loads plus the real. Automatic load

frequency control helps to regulate the MW output of the generator to

maintain the frequency.

It consists of two feedback loop--Primary loop & Secondary loop

system.

Primary loop works for a frequency variation which is a result of

power imbalance in MW. In this loop the output generated voltage

frequency is fed to the governor which raises a signal to control

valves to regulate the steam flow so that output of the generator will

match with fast load fluctuations. This loop matches the initial course

of adjustment of frequency. Response time of this loop is in the order

of 2-20 seconds.

Primary loop - It is a feedback loop system, consists of frequency

sensor, Frequency comparator, Integrator, speed changer, Turbine

governor, Hydraulic amplifier and Valve control mechanism.

Frequency Sensor: It senses the actual frequency of generated voltage

and sends the signal toFrequency comparator.

Frequency comparator: It compares the actual frequency with the

reference frequency and rise the signal ∆f to integrator.

Integrator: It converts frequency signal into speed signal i.e. ∆f to ∆N

and feed it to speed changer.

Speed Changer: According to the signal it rolls low /high side and

activate Turbine governor.

Turbine governor: It is a turbo-generator governor. It raises the signal

to increase or decrease the speed of turbine through Link mechanisms

Link mechanisms: This is a 4-link mechanism, which provides a

movement to Steam valve through Hydraulic amplifier in proportion

to change in speed. It also provides a feedback from the steam valve

movement

Hydraulic amplifier: It comprises a pilot valve and main piston

arrangement. Low power level pilot valve movement is converted

into high power level piston valve movement. This is necessary in

order to open or close the steam valve against high pressure steam.

Valve control mechanism: As per signal valve control mechanism

opens or close the steam /gas valve to increase /decrease the speed i.e.

power output of turbine. Accordingly generator output varies.


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Secondary loop: It comprises Frequency amplifier which receives minor variation signal of frequency and feed the amplified signal to

integrator. Secondary loop works for fine adjustment of the frequency, signal

from the output of the generator is amplified by amplifier and the

amplified signal is fed to integral controls which integrate the frequency error and raise a signal to control the valves which in turn

regulate the steam flow. This loop is sensitive to rapid variations in load frequency. Response time of this loop is in order of one minute.

This ALFC is located in power stations. It is operative only during normal changes in load and frequency. It is unable to provide adequate control during emergency conditions when large MW imbalance occurs.

(c)

Ans.

Explain the different methods of voltage control by using

transformer.

Following are the methods of voltage control in power system. By

using transformers

1. By tap changing transformers.

- Off load tap changing

- On load tap changing

2. By regulating transformers

3. By Booster transformers

4. Auto Transformer tap changing

1) Online tap changing transformer: All transformers are provided with taps on the winding for adjusting

the ratio of transformation. Taps are usually provided on the high

voltage winding to enable a fine control of voltage. Generally the tap

changing can be done any when the transformer is in de-energized

state. However in some cases tap changing is also possible when the

transformer is energized. These transformers make it possible to

maintain a constant voltage level on important buses in the system.

Location: Intermediate distribution Substation

2) Regulating transformer: A special type of transformer designed for small adjustments of

voltage is known regulating transformer. The fig. shows a typical

arrangement to use a regulating transformer for voltage mag control

in a 3 phase ckt. A 3 phase transformer provides on adjustable voltage

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to the primary of the regulating transformer. The secondary of the

regulating transformer are connected in series with the lines. Thus a

voltage magnitude 0volt to the voltage of each phase.

Location: distribution Substation

3) Booster Transformer:

Sometimes it is designed to control the voltage of transmission line at

a point for away from the main transformer. This is conveniently

achieved by the use of booster transformer. The secondary of booster

transformer is connected in series with the line whose voltage is to be

controlled. The primary of this transformer is supplied from a

regulating transformer with on load tap changing gear.

Location: HV &EHV transmission Line

4) Auto transformer tap changing:

Here, a midtapped auto-tranformer or reactor is used. One of the line

is connected its mid-tapping. One end of this transformer is connected

in a series of switches across the odd tapping and the other end is

connected to switches across the even tappings.

(d)

Ans.

How the voltage level can be controlled in power system by

injecting reactive power in the tr. Line?

Reactive power injection method for voltage control:

To keep the receiving end voltage at a specified value, a fixed amount

of VARS must be drawn from the line.

QG =QD bus voltage is maintained at specific value.

As VAR demand QD varies, a local VAR generator must be used as

shown in fig. The VAR balance equation at the receiving end is now.

QG =QD +QC

Any variation QD is are absorbed / injected by the local VAR

generator and QGgenerated by the line remains fixed. This helps to

keep the receiving end voltage constant.

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Following are the equipments used to inject VARs in the system

at different points.

Generation system Excitation control Production of

reactive power

involves increasing

the magnetic field to

raise the generator‟s

terminal voltage. To

increase the

magnetic field,

increase the current

in the field winding.

Absorption of

reactive power is

limited by the

magnetic-flux

pattern in the stator,

which results in

excessive heating of

the stator-end iron,

the core-end heating

limit.

Transmission system Series reactors Capacitors and

inductors in HV and

EHV trans. Line

Static VAR system

Distribution system Shunt reactors Capacitor‟s bank

Synchronous

condenser

Static VAR system

(e)

Ans.

List out the environmental factors that affects the load

forecasting.

List out the environmental factors affecting load forecasting?

Following are Environmental factors that affect load forecasting of

power system.

i.) Time dependent factor

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ii.) Weather dependent factor ( humidity, temperature)

iii.) Wind --wind speed, wind direction, cloud cover, fog

iv.) Random weather change ( storm, sunami, heavy rain, flood)

1M

(f)

Ans. Why social activities are important for power system operation?

Electricity consumers i.e. residential consumer, commercial

consumers, industrial consumers are part of society. Hence their

activities, events affect the power requirement. Following are the

some of the activities that affect the load forecasting of power system.

1) Energy consumption pattern: All 24hr,s of day load on system

varies as consumer has freedom to use electricity whenever they

required without any prior information. Hence daily load curves

differ with the day. Also energy consumption pattern of various

and type of consumer differs. To satisfy all consumers power

generation must be varied with time. So during forecasting of load

these factors must be considered.

2) Holidays/week ends and week days: During power consumption

pattern is nearly same but on weekends / Sundays power

consumption pattern changes. Therefore their impact on load

forecasting cannot be neglected. Public holidays also have

considerable impact on load forecasting. Long weekend‟s creates

more fluctuation in load demand.

3) School /college vacations: Vacation period changes the daily

routine of children and their stay at consumes power for their

activities such as watching TV, playing video games, net surfing,

watching films, etc. So in residential sector more power consumes

by lighting and air-conditioning systems.

4) Festivals and National days: During festivals like Diwali, Dashera,

Christmas, Onam etc. more lightings are used for decoration

purpose. This increases power consumption of residential as well

as commercial sector. Hence they have to be considered for load

forecasting. National days like Independence day, republic day,

Marashtra day etc. all government building are decorated with

lights and more cultural programs were arranged. As power

consumption is of considerable amount, their impact on load has to

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be considered in load forecasting.

5) Emergency conditions and Major accidents: If sudden large

variation power demands, failure of system components, faults

(line-to-line, line-to-ground) in system, causes more imbalances in

power demand and supply. It will put the system in transient

stability condition. Also if major accidents takes place like sunami,

wind storm, earth quack, snow storm, flood etc. may affects the

infrastructure of power system. And so there may be major power

failure. In such situations load forecasting becomes failure.

6) Special events:Labour strike in Industry, political events, VIP

visits also creates large variation in power demand. These events

cannot be neglected

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